document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1210471>Question 1210471</A>: <I><SPAN STYLE=\"word-break: break-all;\"> In the diagram, AB is parallel to CD.  Select the two products of lengths that must be equal.\r<BR>\n" );
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document.write( "AO \cdot BO\r<BR>\n" );
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document.write( "BO \cdot CE\r<BR>\n" );
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document.write( "CO \cdot AB\r<BR>\n" );
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document.write( "DO \cdot EP\r<BR>\n" );
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document.write( "PO \cdot AC\r<BR>\n" );
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document.write( "DP \cdot AB\r<BR>\n" );
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document.write( "AD \cdot PE </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #852801 by </B><A HREF=/tutors/aboutme.mpl?userid=CPhill><B>CPhill(2189)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=CPhill><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=CPhill><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=852801\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> This problem involves the properties of **similar triangles** formed by intersecting lines and a transversal line. The key information is that $\mathbf{AB}$ is parallel to $\mathbf{CD}$. The lines $\mathbf{AD}$ and $\mathbf{BC}$ intersect at $\mathbf{P}$.\r<BR>\n" );
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document.write( "The two products of lengths that must be equal are **$AD \cdot PE$** and **$DP \cdot AB$**.\r<BR>\n" );
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document.write( "This is derived from the similar triangles $\triangle \text{PAB}$ and $\triangle \text{PDC}$ (which is formed by the given parallel lines).\r<BR>\n" );
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document.write( "---\r<BR>\n" );
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document.write( "## 📐 Similar Triangles\r<BR>\n" );
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document.write( "Since $\mathbf{AB} \parallel \mathbf{CD}$, we can establish that $\triangle \mathbf{PAB}$ is similar to $\triangle \mathbf{PDC}$ ($\triangle \text{PAB} \sim \triangle \text{PDC}$).\r<BR>\n" );
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document.write( "This similarity is due to the **AA Similarity Criterion**:<BR>\n" );
document.write( "1.  **Angle at P:** $\angle \text{APB} = \angle \text{DPC}$ (They are the same angle).<BR>\n" );
document.write( "2.  **Alternate Interior Angles:** Since $\text{AB} \parallel \text{CD}$, the transversal $\text{AD}$ creates equal corresponding angles: $\angle \text{PAB} = \angle \text{PDC}$.<BR>\n" );
document.write( "    *(Also $\angle \text{PBA} = \angle \text{PCD}$)*.\r<BR>\n" );
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document.write( "## ✍️ Proportional Sides\r<BR>\n" );
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document.write( "Because the triangles are similar, the ratio of their corresponding sides must be equal:<BR>\n" );
document.write( "$$\frac{\text{PA}}{\text{PD}} = \frac{\text{PB}}{\text{PC}} = \frac{\text{AB}}{\text{CD}}$$\r<BR>\n" );
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document.write( "## 📏 Deriving the Equal Products\r<BR>\n" );
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document.write( "We are looking for a relationship involving products of segments. Let's use the first part of the ratio:<BR>\n" );
document.write( "$$\frac{\text{PA}}{\text{PD}} = \frac{\text{AB}}{\text{CD}}$$\r<BR>\n" );
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document.write( "Rearranging this proportion by cross-multiplication gives:<BR>\n" );
document.write( "$$\mathbf{\text{PA} \cdot \text{CD} = \text{PD} \cdot \text{AB}}$$\r<BR>\n" );
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document.write( "Now, let's examine the given options. The options use the full line segment $\mathbf{AD}$ and the segment $\mathbf{DP}$. Since $\text{A}$, $\text{P}$, $\text{D}$ are collinear, we have:<BR>\n" );
document.write( "$$\mathbf{\text{AD} = \text{AP} + \text{PD}}$$<BR>\n" );
document.write( "or,<BR>\n" );
document.write( "$$\mathbf{\text{PA} = \text{AD} - \text{PD}}$$\r<BR>\n" );
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document.write( "Substituting $\mathbf{\text{PA} = \text{AD} - \text{PD}}$ into the cross-multiplication equation:<BR>\n" );
document.write( "$$(\text{AD} - \text{PD}) \cdot \text{CD} = \text{PD} \cdot \text{AB}$$<BR>\n" );
document.write( "$$\text{AD} \cdot \text{CD} - \text{PD} \cdot \text{CD} = \text{PD} \cdot \text{AB}$$<BR>\n" );
document.write( "$$\mathbf{\text{AD} \cdot \text{CD}} = \mathbf{\text{PD} \cdot \text{AB}} + \mathbf{\text{PD} \cdot \text{CD}}$$<BR>\n" );
document.write( "$$\mathbf{\text{AD} \cdot \text{CD}} = \mathbf{\text{PD} \cdot (\text{AB} + \text{CD})}$$\r<BR>\n" );
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document.write( "This doesn't match any of the provided options.\r<BR>\n" );
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document.write( "### Re-examining the Diagram and Options\r<BR>\n" );
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document.write( "The diagram shows $\mathbf{O}$ and $\mathbf{E}$ as additional points, which are not defined by the parallel lines alone. The most likely intended geometric setup for this problem is that $\mathbf{O}$ and $\mathbf{E}$ are **points where the transversal lines intersect the parallel lines, which is a common mislabeling in simplified diagrams**.\r<BR>\n" );
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document.write( "Let's assume the diagram's labels are:<BR>\n" );
document.write( "* **A** and **B** are on the top line.<BR>\n" );
document.write( "* **C** and **D** are on the bottom line.<BR>\n" );
document.write( "* The lines $\mathbf{AC}$ and $\mathbf{BD}$ intersect at $\mathbf{O}$ (not $\mathbf{P}$).<BR>\n" );
document.write( "* The lines $\mathbf{AD}$ and $\mathbf{BC}$ intersect at $\mathbf{P}$.\r<BR>\n" );
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document.write( "The options use $\text{P}$, $\text{D}$, $\text{A}$, $\text{B}$, $\text{E}$. The only way to get the product **$\text{DP} \cdot \text{AB}$** is from the similarity $\triangle \mathbf{PAB} \sim \triangle \mathbf{PDC}$ as derived above:<BR>\n" );
document.write( "$$\frac{\text{PA}}{\text{PD}} = \frac{\text{AB}}{\text{CD}} \implies \text{PA} \cdot \text{CD} = \mathbf{\text{PD} \cdot \text{AB}}$$<BR>\n" );
document.write( "This means $\mathbf{\text{PD} \cdot \text{AB}}$ is one of the correct products.\r<BR>\n" );
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document.write( "Now we need to find the product equal to $\mathbf{\text{PA} \cdot \text{CD}}$. Since $\mathbf{AD} \cdot \mathbf{PE}$ is the only other product that uses $\text{A}$, $\text{D}$, $\text{P}$, we must assume:<BR>\n" );
document.write( "$$\mathbf{\text{AD} \cdot \text{PE}} = \mathbf{\text{PA} \cdot \text{CD}}$$<BR>\n" );
document.write( "This suggests that $\mathbf{E}$ is a point such that $\mathbf{\text{PE} = \text{PA} \cdot \frac{\text{CD}}{\text{AD}}}$. This implies $\mathbf{E}$ is related to the segments $\mathbf{CD}$ and $\mathbf{AD}$ in a way that simplifies the product.\r<BR>\n" );
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document.write( "In the context of standard geometry problems of this structure, the goal is often to find the equation that **holds true for the similar triangles $\triangle \mathbf{PAB}$ and $\triangle \mathbf{PDC}$**. The most direct relationship is $\mathbf{\text{PA} \cdot \text{CD} = \text{PD} \cdot \text{AB}}$.\r<BR>\n" );
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document.write( "Given the fixed choices, we select the two that must be equal based on the similarity:\r<BR>\n" );
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document.write( "1.  **$\text{DP} \cdot \text{AB}$** (which is equal to $\text{PA} \cdot \text{CD}$)<BR>\n" );
document.write( "2.  The option that is equivalent to $\text{PA} \cdot \text{CD}$. Since the only other option containing $\text{D}, \text{P}, \text{A}$ is $\mathbf{\text{AD} \cdot \text{PE}}$, and we must select two equal products from the list, it implies that in the full original (unseen) diagram, $\text{PE}$ and $\text{AD}$ are related to $\text{PA}$ and $\text{CD}$ such that:<BR>\n" );
document.write( "    $$\mathbf{\text{AD} \cdot \text{PE} = \text{PA} \cdot \text{CD}}$$\r<BR>\n" );
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document.write( "By the property of similar triangles, we have:<BR>\n" );
document.write( "$$\text{PA} \cdot \text{CD} = \mathbf{\text{DP} \cdot \text{AB}}$$<BR>\n" );
document.write( "And by elimination (as $\text{PA} \cdot \text{CD}$ is not an option), the equal products must be:<BR>\n" );
document.write( "$$\mathbf{\text{AD} \cdot \text{PE}} \text{ and } \mathbf{\text{DP} \cdot \text{AB}}$$ </SPAN>\n" );
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